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melisa1 [442]
3 years ago
11

Galileo attempted to measure the speed of light by measuring the time elapsed between his opening a lantern and his seeing the l

ight return from his assistant's lantern. What distance, d, must separate Galileo and his assistant in order for the human reaction time, Δt = 0.2 s, to introduce no more than a 14 % error in the speed of light? Answer in meters.
Physics
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

The distance is 2.58\times10^{7}\ m.

Explanation:

Given that,

Time \Delta t = 0.2\ s

The velocity is no more than a 14 % error in the speed of light.

So,

Velocity v= c\times 86\%

We need to calculate the distance

Using formula of speed

v = \dfrac{d}{t}

d = v\times t

Where, v = speed

d = distance

t = time

Put the value into the formula

d = 3\times10^{8}\times\dfrac{86}{100}\times0.2

d=516\times10^{5}\ m

We know that,

The one side distance d' is

d'=\dfrac{d}{2}

d'=\dfrac{516\times10^{5}}{2}

d'=2.58\times10^{7}\ m

Hence, The distance is 2.58\times10^{7}\ m.

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Answer:

5m/s²

Explanation:

Given parameters:

Mass of wagon  = 10kg

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Frictional force = 20N

Unknown:

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Solution:

Frictional force is a force that opposes motion.

The net force is given as:

   Net force  = mass x acceleration

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     70 - 20  = 10 x acceleration

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A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and
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<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.

<h3>What is a momentum?</h3>
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  • It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

7 0
3 years ago
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